欧拉准则(检查模p下的平方根是否存在)

给定一个数字n和一个质数p，求模p下n的平方根是否存在。如果(x * x)％p = n％p，则数字x是n在模p下的平方根。

例子 ：

Input:   n = 2, p = 5
Output:  false
There doesn't exist a number x such that 
(x*x)%5 is 2

Input:   n = 2, p = 7
Output:  true
There exists a number x such that (x*x)%7 is
2.  The number is 3.

朴素的方法是尝试x介于2到p-1之间的每个数字x。对于每个x，检查(x * x)％p是否等于n％p。

C++

// A Simple C++ program to check if square root of a number
// under modulo p exists or not
#include
using namespace std;
 
// Returns true if square root of n under modulo p exists
bool squareRootExists(int n, int p)
{
    n = n%p;
 
    // One by one check all numbers from 2 to p-1
    for (int x=2; x

---

Java

// A Simple Java program to check if square
// root of a number under modulo p exists or not
class GFG
{
     
    // Returns true if square root of n under
    // modulo p exists
    static boolean squareRootExists(int n, int p)
    {
        n = n % p;
     
        // One by one check all numbers from 2
        // to p-1
        for (int x = 2; x < p; x++)
            if ((x * x) % p == n)
                return true;
                 
        return false;
    }
     
    // Driver program to test
    public static void main (String[] args)
    {
         
        int p = 7;
        int n = 2;
         
        if(squareRootExists(n, p))
            System.out.print("Yes");
        else
            System.out.print("No"); 
    }
}
 
// This code is contributed by Anant Agarwal.

Python3

# A Simple Python 3 program to
# check if square root of a number
# under modulo p exists or not
 
# Returns true if square root of
# n under modulo p exists
def squareRootExists(n, p):
    n = n % p
 
    # One by one check all numbers
    # from 2 to p-1
    for x in range(2, p, 1):
        if ((x * x) % p == n):
            return True
    return False
 
# Driver Code
if __name__ == '__main__':
    p = 7
    n = 2
    if(squareRootExists(n, p) == True):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by
# Surendra_Gangwar

C#

// A Simple C# program to check
// if square root of a number
// under modulo p exists or not
using System;
 
class GFG
{
     
    // Returns true if square root of 
    // n under modulo p exists
    static bool squareRootExists(int n,
                                 int p)
    {
        n = n % p;
     
        // One by one check all numbers
        // from 2 to p-1
        for (int x = 2; x < p; x++)
            if ((x * x) % p == n)
                return true;
                 
        return false;
    }
     
    // Driver code
    public static void Main ()
    {
         
        int p = 7;
        int n = 2;
         
        if(squareRootExists(n, p))
            Console.WriteLine("Yes");
        else
            Console.WriteLine("No");
    }
}
 
// This code is contributed by Sam007.

Javascript

 
// A Simple Javascript program to check
// if square root of a number
// under modulo p exists or not
 
// Returns true if square root
// of n under modulo p exists
function squareRootExists(n, p)
{
    n = n % p;
     
    // One by one check all
    // numbers from 2 to p-1
    for(let x = 2; x < p; x++)
        if ((x * x) % p == n)
            return true;
             
    return false;
}
 
// Driver Code
let p = 7;
let n = 2;
 
if (squareRootExists(n, p) === true)
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by _saurabh_jaiswal
 

输出 ：

Yes

该方法的时间复杂度为O(p)。

这个问题有一个基于欧拉准则的直接解决方案。
欧拉准则指出

Square root of n under modulo p exists if and only if
n(p-1)/2 % p = 1

Here square root of n exists means is, there exist
an integer x such that (x * x) % p = 1

以下是基于上述标准的实现。有关幂函数，请参见模幂。

C++

// C++ program to check if square root of a number
// under modulo p exists or not
#include
using namespace std;
 
// Utility function to do modular exponentiation.
// It returns (x^y) % p.
int power(int x, int y, int p)
{
    int res = 1;     // Initialize result
    x = x % p; // Update x if it is more than or
               // equal to p
 
    while (y > 0)
    {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res*x) % p;
 
        // y must be even now
        y = y>>1; // y = y/2
        x = (x*x) % p;
    }
    return res;
}
 
// Returns true if there exists an integer x such
// that (x*x)%p = n%p
bool squareRootExists(int n, int p)
{
    // Check for Euler's criterion that is
    // [n ^ ((p-1)/2)] % p is 1 or not.
    if (power(n, (p-1)/2, p) == 1)
       return true;
 
    return false;
}
 
// Driver program to test
int main()
{
   int p = 7;
   int n = 2;
   squareRootExists(n, p)? cout << "Yes": cout << "No";
   return 0;
}

Java

// Java program to check if
// square root of a number
// under modulo p exists or not
import java.io.*;
 
class GFG
{
     
// Utility function to do
// modular exponentiation.
// It returns (x^y) % p.
static int power(int x, int y, int p)
{
    int res = 1;     // Initialize result
    x = x % p; // Update x if it is more
               // than or equal to p
 
    while (y > 0)
    {
        // If y is odd, multiply
        // x with result
        if ((y & 1) == 0)
            res = (res * x) % p;
 
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
 
// Returns true if there
// exists an integer x such
// that (x*x)%p = n%p
static boolean squareRootExists(int n,
                                int p)
{
    // Check for Euler's criterion
    // that is [n ^ ((p-1)/2)] % p
    // is 1 or not.
    if (power(n, (p - 1) / 2, p) == 1)
    return true;
 
    return false;
}
 
// Driver Code
public static void main (String[] args)
{
    int p = 7;
    int n = 2;
    if(squareRootExists(n, p))
        System.out.println ("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by ajit

Python3

# Python3 program to check if square root
# of a number under modulo p exists or not
 
# Utility function to do modular
# exponentiation. It returns (x^y) % p.
def power(x, y, p):
    res = 1 # Initialize result
    x = x % p
     
    # Update x if it is more than
    # or equal to p
    while (y > 0):
         
        # If y is odd, multiply
        # x with result
        if (y & 1):
            res = (res * x) % p
             
        # y must be even now
        y = y >> 1 # y = y/2
        x = (x * x) % p
    return res
 
# Returns true if there exists an integer
# x such that (x*x)%p = n%p
def squareRootExists(n, p):
     
    # Check for Euler's criterion that is
    # [n ^ ((p-1)/2)] % p is 1 or not.
    if (power(n, (int)((p - 1) / 2), p) == 1):
        return True
    return False
 
# Driver Code
p = 7
n = 2
if(squareRootExists(n, p) == True):
    print("Yes")
else:
    print("No")
 
# This code is contributed by Rajput-Ji

C＃

// C# program to check if
// square root of a number
// under modulo p exists or not
using System;
 
class GFG
{
     
// Utility function to do
// modular exponentiation.
// It returns (x^y) % p.
static int power(int x,
                 int y, int p)
{
    int res = 1;// Initialize result
    x = x % p;  // Update x if it is more
                // than or equal to p
 
    while (y > 0)
    {
        // If y is odd, multiply
        // x with result
        if ((y & 1) == 0)
            res = (res * x) % p;
 
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
 
// Returns true if there
// exists an integer x such
// that (x*x)%p = n%p
static bool squareRootExists(int n,
                             int p)
{
    // Check for Euler's criterion
    // that is [n ^ ((p-1)/2)] % p
    // is 1 or not.
    if (power(n, (p - 1) / 2, p) == 1)
    return true;
 
    return false;
}
 
// Driver Code
static public void Main ()
{
    int p = 7;
    int n = 2;
    if(squareRootExists(n, p))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by m_kit

的PHP

 0)
    {
        // If y is odd, multiply
        // x with result
        if ($y & 1)
            $res = ($res *
                    $x) % $p;
 
        // y must be even now
        $y = $y >> 1; // y = y/2
        $x = ($x * $x) % $p;
    }
    return $res;
}
 
// Returns true if there
// exists an integer x such
// that (x*x)%p = n%p
function squareRootExists($n, $p)
{
    // Check for Euler's criterion
    // that is [n ^ ((p-1)/2)] % p
    // is 1 or not.
    if (power($n, (int)(($p - 1) / 2),
                         $p) == 1)
    return true;
 
    return false;
}
 
// Driver Code
$p = 7;
$n = 2;
 
if(squareRootExists($n, $p) == true)
    echo "Yes";
else
    echo "No";
 
// This code is contributed by ajit
?>

Java脚本

 
    // Javascript program to check if
    // square root of a number
    // under modulo p exists or not
     
    // Utility function to do
    // modular exponentiation.
    // It returns (x^y) % p.
    function power(x, y, p)
    {
        let res = 1;// Initialize result
        x = x % p;  // Update x if it is more
                    // than or equal to p
 
        while (y > 0)
        {
            // If y is odd, multiply
            // x with result
            if ((y & 1) == 0)
                res = (res * x) % p;
 
            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }
        return res;
    }
 
    // Returns true if there
    // exists an integer x such
    // that (x*x)%p = n%p
    function squareRootExists(n, p)
    {
        // Check for Euler's criterion
        // that is [n ^ ((p-1)/2)] % p
        // is 1 or not.
        if (power(n, (p - 1) / 2, p) == 1)
            return true;
 
        return false;
    }
     
    let p = 7;
    let n = 2;
    if(squareRootExists(n, p))
        document.write("Yes");
    else
        document.write("No");
 

输出 ：

Yes

这是如何运作的?

If p is a prime, then it must be an odd number and (p-1) 
must be an even, i.e., (p-1)/2 must be an integer.

Suppose a square root of n under modulo p exists, then
there must exist an integer x such that,
      x2 % p = n % p 
or, 
     x2 ? n mod p

Raising both sides to power (p-1)/2,
      (x2)(p-1)/2 ? n(p-1)/2 mod p           
      xp-1 ? n(p-1)/2 mod p

Since p is a prime, from Fermet's theorem, we can say that 
   xp-1 ? 1 mod p

Therefore,
  n(p-1)/2 ? 1 mod p